2 Answers
2

You're on the right track, but it would be wise to say $c=jm+d$, to use a different constant from $k$. Then
$$
ac=(km+b)(jm+d)=kjm^2+bjm+dkm+bd=(kjm+bj+dk)m+bd.
$$
Since $(kjm+bj+dk)m$ is a multiple of $m$, you've shown $ac\equiv bd\pmod{m}$.